Question

If f(x)= x+1^-1 and g(x)=x-2 , what is the domain of f(x) divided by g(x)

Answer 1

what do you think

Answer 2

honestly I don't know, I want to say all values of x but im not sure :(

Answer 3

Can you write out \(\dfrac{f(x)}{g(x)}\)

Answer 4

no I don't know how sorry

Answer 5

can you guys help me pleaseeee

Answer 6

x+1^-1, So your saying the whole quantity by -1, or just "1"

Answer 7

\[\frac{ x+1 }{ x-2 }\]

Answer 8

yeah the exponent is -1

Answer 9

its the inverse \((x+1)^{-1}=\dfrac{1}{x+1}\)

Answer 10

no

Answer 11

The exponent is only on 1 not (x+1)

Answer 12

So you have \(\dfrac{(x+1)^{-1}}{x-2}=\dfrac{1}{(x+1)(x-2)}\).

Now what numbers force us to divide by \(0\)?

Answer 13

\[Did she mean (x+1)^{-1}\]

Answer 14

The zeroes of (x+1)(x-2)

Answer 15

yes. there is no reason to raise something to 1

Answer 16

Also known as (x+1)=0, (x-2)=0, Solve for X on both of them @idgm_idontgetmath

Answer 17

x=-1,2

Answer 18

please let them figure something out man. you are not helping if you give thm everything....

Answer 19

But im trying to show him to solve zeroes??????

Answer 20

Not you.

Answer 21

@carolinar7

Answer 22

@vera_ewing

Answer 23

@juanpabloJR

Answer 24

@campbell_st can you help?

Answer 25

ok... here is what I'll do, if you are happy that

\[\frac{f(x)}{g(x)} = \frac{1}{(x +1)(x-2)}\]

you need to solve

x + 1 = 0 and x - 2 = 0

can you do that

Answer 26

so here is an image of f(x)/g(x)

Answer 27

the curve continues to the left.. to- infinity

and to the right to infinity

however the solutions to

x + 1 = 0 and x -2 = 0 create vertical asymptotes... this is where the cruve doesn't exist

but you can see that the curve exists between the asymptotes

so there are 3 regions where the curve exists or can be drawn...

Answer 28

so the regions start with

- infinity to the 1st asymptote

between the asymptotes

and the right asymptote to infinity... this will be the domain

Answer 29

If im understanding right the solution is B then

Answer 30

that's correct

Answer 31

awww thanks